most recent 30 from http://mathoverflow.net 2013-06-19T07:23:54Z http://mathoverflow.net/feeds/question/61116 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61116/lie-algabra-of-symmetric-group Dr Shello 2011-04-09T05:25:44Z 2011-04-12T21:04:52Z

<p>It's easy to see that the descending central series of a group induces a graded Lie algebra .(see for example Serre's Harvard lectures or Magnus-Solitar book). I think in general this can be complicated, but this should be well-known:</p> <blockquote> <p>What is the structure of this Lie algebra for the symmetric group?</p> </blockquote>

http://mathoverflow.net/questions/61116/lie-algabra-of-symmetric-group/61144#61144 S. Carnahan 2011-04-09T15:42:01Z 2011-04-09T15:42:01Z

<p>For any $n>1$, the lower central series for the symmetric group is $S_n > A_n > A_n > A_n > \cdots$, so the Lie ring formed by the sum of successive quotients is the group $\mathbb{Z}/2\mathbb{Z}$, equipped with the Lie bracket that is identically zero.</p> <p>If you want to gain intuition for this construction with finite groups, I suggest you consider nilpotent groups, since their lower central series actually reach the trivial group. For example, many $p$-groups will yield nonabelian Lie algebras over $\mathbb{F}_p$.</p>

http://mathoverflow.net/questions/61116/lie-algabra-of-symmetric-group/61470#61470 James 2011-04-12T21:04:52Z 2011-04-12T21:04:52Z

<p>For a well-known infinite example, there is a result, due to Labute ("On the descending central series of groups with a single defining relation", J. Algebra 14 (1970), 16--23) which asserts that the Lie ring associated to a one-relator group can be presented as a Lie ring with a single defining relator.</p>